First of all, grep searches through the contents of files, it will not search for file names. For that you want find. It does not have an option to set the filename length but you can parse it's output to get what you want. Here are some examples that I run in a directory containing the following files:

So, to find the files with a length of X, we will need to remove the path and match only the characters after the last /:

$ find . -type f | grep -P '/.{3}$'

The sed command just removes the ./. The -P flag activates Perl Compatible Regular Expressions which are needed for {n} which is a PCRE thing and the $ means "match the end of the string".

You could also simply do this:

find . -type f | grep -P '/...$'

That's fine for small numbers typing 15 dots to match filenames of length 15 is not very efficient.

Finally, if you want to ignore the extension and match only the file's name, do this (as @slm suggested):

find . -type f | grep -P '/.{1}\.'

However, this will also find files like a.bo.go. This is better if your filenames can contain more than one . :

find . -type f | grep -P '/.{1}\.[^.]+$'

NOTE: The solution above assumes that you have relatively sane filenames that don't contain new line characters etc. It will also count not ignore spaces in filenames when calculating the length which might not be what you want. If either of these is a problem, let me know and I'll update my answer.